Series AC circuit element parameters

Step 1: Bring both quantities to the same form.

$$i=20\sqrt2\cos(314t-45^\circ)=20\sqrt2\sin(314t+45^\circ)$$ $$v=400\sqrt2\sin(314t+40^\circ)$$

RMS values: $V=400$ V, $I=20$ A. Phase angle of current is $+45^\circ$ and of voltage $+40^\circ$, so the current leads the voltage by

$$\phi = 45^\circ-40^\circ = 5^\circ$$

A leading current means the circuit is a series R–C circuit.

Step 2: Impedance and parameters.

$$\begin{aligned} Z &= \frac{V}{I}=\frac{400}{20}=20\ \Omega\\ R &= Z\cos\phi = 20\cos5^\circ = 19.92\ \Omega\\ X_C &= Z\sin\phi = 20\sin5^\circ = 1.743\ \Omega\\ C &= \frac{1}{\omega X_C}=\frac{1}{314\times1.743}=1827\ \mu\text{F} \end{aligned}$$

Step 3: Voltage drops.

$$\begin{aligned} V_R &= IR = 20\times19.92 = 398.5\ \text{V}\\ V_C &= IX_C = 20\times1.743 = 34.86\ \text{V} \end{aligned}$$

Check: $\sqrt{398.5^2+34.86^2}=400$ V.

Step 4: Power factor and power.

$$\cos\phi=\cos5^\circ=0.996\ \text{(leading)}$$ $$P = VI\cos\phi = 400\times20\times0.996 = 7970\ \text{W}\ (=I^2R)$$

Answer: $R=19.92\ \Omega$, $C=1827\ \mu$F ($X_C=1.743\ \Omega$); $V_R=398.5$ V, $V_C=34.86$ V; pf = 0.996 leading; $P=7.97$ kW.